Reactions of the halide ions · 卤离子的反应
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| halide/ˈhælaɪd/ | 卤离子 | lǔ lí zi |
| reducing agent/rɪˈdjuːsɪŋ ˈeɪdʒənt/ | 还原剂 | huán yuán jì |
| precipitate/prɪˈsɪpɪteɪt/ | 沉淀 | chén diàn |
| silver nitrate/ˈsɪlvə ˈnaɪtreɪt/ | 硝酸银 | xiāo suān yín |
Spotting a halide 卤离子
- A halide ion can give away an electron — so it is a reducing agent 还原剂.
- This power increases down the group.
- Two key tests identify which halide is present.
识别一个卤离子
- 一个卤离子(halide ion)能给出一个电子——所以它是一个还原剂(reducing agent)。
- 这个能力沿族向下增加。
- 两个关键检验识别哪个卤离子存在。
Halide ion test lab · 卤离子检验实验室
Match halide ion evidence to the ion present. · 把卤离子证据匹配到存在的离子。
The reducing power of the halide ions increases down the group because: · 卤离子的还原能力沿族向下增加,因为:
Reducing means losing an electron; bigger ions lose theirs more easily, so reducing power rises down the group. · 还原意味着失去一个电子;更大的离子更容易失去它们的,所以还原能力沿族向下上升。
Silver nitrate solution is used to test for ______ ions. · 硝酸银溶液用于检验______离子。
Different halides give different coloured precipitates. · 不同的卤离子给出不同颜色的沉淀。
Reducing power
- A halide ion ($\text{Cl}^{-}$, $\text{Br}^{-}$, $\text{I}^{-}$) acts as a reducing agent (it loses an electron).
- This power increases down the group, because a larger ion holds its outer electron less tightly.
Silver halide precipitate 沉淀 colours and their solubility in ammonia identify the halide
还原能力
- 一个卤离子($\text{Cl}^{-}$、$\text{Br}^{-}$、$\text{I}^{-}$)作为一个还原剂(它失去一个电子)。
- 这个能力沿族向下增加,因为一个更大的离子把它的外层电子抓得不那么紧。
银卤化物沉淀的颜色和它们在氨中的溶解度识别卤离子
Match each halide to its silver halide precipitate colour. · 把每个卤离子匹配到它的银卤化物沉淀颜色。
AgCl is white, AgBr cream, AgI yellow; ammonia solubility confirms (Cl dissolves in dilute, I insoluble). · AgCl 是白色,AgBr 奶油色,AgI 黄色;氨溶解度确认(Cl 溶于稀氨,I 不溶)。
Silver iodide (AgI) is: · 碘化银(AgI)是:
AgI is yellow and does not dissolve in ammonia, distinguishing it from AgCl (white) and AgBr (cream). · AgI 是黄色且不溶于氨,把它与 AgCl(白色)和 AgBr(奶油色)区分开。
Silver nitrate 硝酸银 test
Add silver nitrate, then ammonia, and read the precipitate:
| Halide | Precipitate | In ammonia |
|---|---|---|
| $\text{Cl}^{-}$ | white | dissolves in dilute ammonia |
| $\text{Br}^{-}$ | cream | dissolves in concentrated ammonia |
| $\text{I}^{-}$ | yellow | insoluble |
硝酸银检验
加入硝酸银,然后氨,读取沉淀:
| 卤离子 | 沉淀 | 在氨中 |
|---|---|---|
| $\text{Cl}^{-}$ | 白色 | 溶于稀氨 |
| $\text{Br}^{-}$ | 奶油色 | 溶于浓氨 |
| $\text{I}^{-}$ | 黄色 | 不溶 |
With concentrated sulfuric acid, iodide gives I₂ plus H₂S because: · 用浓硫酸,碘离子给出 I₂ 加 H₂S,因为:
I⁻ is the strongest reducer, so it reduces the sulfuric acid all the way to H₂S (and forms I₂); Cl⁻ only gives HCl. · I⁻ 是最强的还原剂,所以它把硫酸一路还原到 H₂S(并形成 I₂);Cl⁻ 只给出 HCl。
With concentrated sulfuric acid
- All give the hydrogen halide first; the lower halides are then oxidised (they are stronger reducing agents):
- chloride: only $\text{HCl}$ (no redox).
- bromide: also brown $\text{Br}_2$ and $\text{SO}_2$.
- iodide: $\text{I}_2$ plus smelly $\text{H}_2\text{S}$ and $\text{SO}_2$ ($\text{I}^{-}$ is the strongest reducing agent).
与浓硫酸
- 全都先给出卤化氢;然后较低的卤离子被氧化(它们是更强的还原剂):
- 氯离子:只有 $\text{HCl}$(无氧化还原)。
- 溴离子:还有棕色 $\text{Br}_2$ 和 $\text{SO}_2$。
- 碘离子:$\text{I}_2$ 加上有臭味的 $\text{H}_2\text{S}$ 和 $\text{SO}_2$($\text{I}^{-}$ 是最强的还原剂)。
You've got it
- halide ions are reducing agents; power increases down the group (larger ion, electron held loosely)
- silver nitrate test: AgCl white, AgBr cream, AgI yellow (confirm with ammonia solubility)
- conc. H₂SO₄: Cl⁻ → HCl only; Br⁻ → Br₂ + SO₂; I⁻ → I₂ + H₂S + SO₂ (strongest reducer)
你掌握了
- 卤离子是还原剂;能力沿族向下增加(更大的离子,电子抓得松)
- 硝酸银检验:AgCl 白色,AgBr 奶油色,AgI 黄色(用氨溶解度确认)
- 浓 H₂SO₄:Cl⁻ → 只有 HCl;Br⁻ → Br₂ + SO₂;I⁻ → I₂ + H₂S + SO₂(最强的还原剂)